2cos^2(x-3п/2)-sin(x-п) =0 [(5п/2) ; 4п ]

2sin²x+sinx=0sinx(2sinx+1)=0sinx=0⇒x=πn,n∈Z5π/2≤πn≤4π5/2≤n≤4n=3⇒x=3πn=4⇒x=4πsinx=-1/2⇒x=(-1)^(k+1)*π/6+πk,k∈Zk=2⇒x=-π/6+2π=11π/6∉[5π/2;4π]k=3⇒x=π/6+3π=19π/6k=4⇒x=-π/6+4π=23π/6k=5⇒x=π/6+5π=31π/6∉[5π/2;4π]